Question: $ D = \left[\begin{array}{rr}4 & -1 \\ 5 & -2\end{array}\right]$ $ F = \left[\begin{array}{rrr}2 & 3 & -2 \\ 4 & -2 & 2\end{array}\right]$ What is $ D F$ ?
Explanation: Because $ D$ has dimensions $(2\times2)$ and $ F$ has dimensions $(2\times3)$ , the answer matrix will have dimensions $(2\times3)$ $ D F = \left[\begin{array}{rr}{4} & {-1} \\ {5} & {-2}\end{array}\right] \left[\begin{array}{rrr}{2} & \color{#DF0030}{3} & \color{#9D38BD}{-2} \\ {4} & \color{#DF0030}{-2} & \color{#9D38BD}{2}\end{array}\right] = \left[\begin{array}{rrr}? & ? & ? \\ ? & ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rrr}{4}\cdot{2}+{-1}\cdot{4} & ? & ? \\ ? & ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rrr}{4}\cdot{2}+{-1}\cdot{4} & ? & ? \\ {5}\cdot{2}+{-2}\cdot{4} & ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rrr}{4}\cdot{2}+{-1}\cdot{4} & {4}\cdot\color{#DF0030}{3}+{-1}\cdot\color{#DF0030}{-2} & ? \\ {5}\cdot{2}+{-2}\cdot{4} & ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rrr}{4}\cdot{2}+{-1}\cdot{4} & {4}\cdot\color{#DF0030}{3}+{-1}\cdot\color{#DF0030}{-2} & {4}\cdot\color{#9D38BD}{-2}+{-1}\cdot\color{#9D38BD}{2} \\ {5}\cdot{2}+{-2}\cdot{4} & {5}\cdot\color{#DF0030}{3}+{-2}\cdot\color{#DF0030}{-2} & {5}\cdot\color{#9D38BD}{-2}+{-2}\cdot\color{#9D38BD}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rrr}4 & 14 & -10 \\ 2 & 19 & -14\end{array}\right] $